Question

OAsk Your Teacher 2. +1/3 polnts | Previous Answers SF6 10.P.015 My Notes A brass ring of diameter 10 cm at 18.0°C is heated and slipped over an aluminum rod with a diameter of 10.01 cm at 18.0°C. Assume that the average coefficients of linear expansion are constant (a) To what temperature must this combination be cooled to separate them? PC Is this temperature attainable? yes no (b) If the aluminum rod were 10.03 cm in diameter, what would be the required temperature? oC

Answer 1

Given : L_Al = 10.01cm ; L_Brass = 10 cm

Constant : $\alpha$_Al = 24×10^-6 ; $\alpha$_Brass = 19×10^-6

Solution :

(a)

Using the equation :

L_Al(1+$\alpha$_Al$\Delta$T) = L_Brass(1+$\alpha$_Brass$\Delta$T)

$\Delta$T = [L_Al -L_Brass] ÷ [ L_Brass$\alpha$_Brass - L_Al$\alpha$_Al]

= [10.01 - 10] ÷ [(10)(19×10^-6) - (10.01)(24×10^-6)]

= -199°c

So T = -199°c + 18°c = 181°c

Yes This is attainable because it is more that 0 K

(b)

L_Al = 10.03 cm

i.e.

$\Delta$T = [10.03 - 10] ÷ [ (10)(19×10^-6) - (10.03)(24×10^-6)]

= -591.48°c

So , T = -591.48°c + 18°c = -573.48°c

This can not be reached as it is less than 0 K.